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In three-phase circuit analysis, determining the proper source voltage for each phase is crucial. Each phase voltage must be calculated based on the given sequence and known phase voltages. In our example, the b-phase voltage is known as \(20 \angle -90^{\circ}\text{ V}\) and the sequence is acb. This sequence implies an anti-clockwise arrangement. To find the a-phase voltage, we introduce a phase shift of \(-120^{\circ}\) from b-phase. Thus, the calculation is \(V_a = 20 \angle -90^{\circ} + 120^{\circ} = 20 \angle 30^{\circ} \text{ V}\). Understanding these sequence relationships is vital. Adding or subtracting the correct phase angle ensures that everything remains balanced, a key feature in three-phase systems.

Impedance plays a critical role in understanding how currents and voltages behave in AC circuits. It represents the total opposition a circuit offers to the flow of alternating current and is comprised of two components: resistance (R) and reactance (X).

• Resistance is the real part and does not change with frequency.
• Reactance is the imaginary part and depends on frequency, indicating how inductors and capacitors oppose AC.

To find the total impedance in a circuit, sum the resistances and reactances of the components.
In our exercise, the line and load impedances are added to form a total impedance \(Z_{tot} = 118 - j96 \Omega\). Understanding impedance is crucial for determining how voltages and currents interact within the circuit.

Ohm's Law is a fundamental principle that links current (I), voltage (V), and impedance (Z) in AC circuits. It's expressed as \(I = \frac{V}{Z}\). This formula helps calculate unknown values when two of the parameters are known.
In the given problem, Ohm's Law is used to find the a-phase line current, \(I_a\). With the source voltage \(V_a = 20 \angle 30^{\circ} \text{ V}\) and total impedance \(Z_{tot} = 118 - j96 \Omega\), we use Ohm's Law to compute:
\[ I_a = \frac{20 \angle 30^{\circ}}{118 - j96} \]First, calculate the magnitude \(|Z_{tot}| = 153 \Omega\) and angle \(\theta = \text{atan} \left( \frac{-96}{118} \right) \approx -39.72^{\circ}\). Finally, \(I_a = 0.13 \angle 69.72^{\circ} \text{ A}\). This calculation underscores the relationship between voltage, impedance, and current passing through the circuit.

The Voltage Divider Rule is a handy method for determining the voltage across a particular component in series within a circuit.
It states that the voltage across a load can be found using the current flowing through it and its impedance.
In the context of our exercise, the a-phase line voltage \(V_{La}\) is calculated as follows:
\[ V_{La} = I_a \cdot Z_{load} \]Given the line current \(I_a = 0.13 \angle 69.72^{\circ}\text{ A}\) and load impedance \(Z_{load} = 117 - j99 \Omega\), calculate
\[ |Z_{load}| = \sqrt{117^2 + (-99)^2} = 153 \Omega \]and the angle \(\theta = \tan^{-1}\left(\frac{-99}{117}\right) \approx -40.72^{\circ}\). Finally, compute\( V_{La} \approx 19.9 \angle 29^{\circ} \text{ V}\).The Voltage Divider Rule simplifies understanding of how voltage varies across components in a complex circuit.